# Integrate the function$\frac{x^3\sin\bigg(\tan^{-1}x^4\bigg)}{1+x^8}$

$\begin{array}{1 1}\large \frac{-\cos(\tan^{-1}x^4)}{4}+c \\ \large \frac{\cos(\tan^{-1}x^4)}{4}+c \\ \large \frac{-\sin (\tan^{-1}x^4)}{4}+c \\ \large \frac{\sin (\tan^{-1}x^4)}{4}+c \end{array}$

Toolbox:
• (i)Method of substitution :
• Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
• Consider $I=\int f(x)dx.$
• Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
• $\Rightarrow$dx=g'(t)dt.
• Thus $I=\int f(g(t).g'(t))dt.$
• (ii)$\int \frac{1}{x}dx=log x+c.$
Given $I=\int \frac{x^3\sin (\tan^{-1}x^4)}{1+x^8}dx$.

Put $\tan^{-1}x^4=t$

$\frac{1}{1+x^8}4x^3dx=dt.$

$\Rightarrow \frac{x^3}{1+x^8}dx=\frac{dt}{4}.$

Now substituting of x and dx we get,

$I=\int \frac{dt}{4}.t\sin t=\frac{1}{4}\int\sin t.dt.$

On integrating we get,

$\;\;\;=\frac{-1}{4}\cos(t)+c.$

Substituting back for t we get,

$\int \frac{x^3\sin (\tan^{-1}x^4)}{1+x^8}dx=\frac{-1}{4}\cos(\tan^{-1}x^4)+c.$